help-0

[raw bone]

cant figure out b or c

,ask d/dydx [xy (x^2-y^2)/(x^2+y^2)]

[ocean seal]

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[raw bone]

,ask df/dydx [xy (x^2-y^2)/(x^2+y^2)]

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Weppman#2037

[raw bone]

,ask d/dxdy [xy (x^2-y^2)/(x^2+y^2)]

[ocean seal]

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Requested by Weppman#2037

[raw bone]

<@&286206848099549185>

[olive loom]

you have $\frac{\partial f}{\partial y}$ at (x, 0) as x and $\frac{\partial f}{\partial x}$ at (0, y) = -y
for b, solve $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)$ and $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$
both will turn out to be constants and not the same

[ocean seal]

Alpha Kappa

[raw bone]

i dont understands how to solve for b

[olive loom]

$\frac{\partial f}{\partial y}=x$

[ocean seal]

Alpha Kappa

[olive loom]

you got this?

[raw bone]

oh

kinda

not really

[olive loom]

have you studied partial differentiation?

[raw bone]

yeah but i dont understand it well

kinda why im here

[olive loom]

i see

in partial differentiation

if you have a function f which is a function of both x and why ie,
f(x, y)
and you are asked to take the partial derivative of f wrt to x
then you have to assume all other variables as a constant

[raw bone]

i understand this

i was able to do the first question

ill post a pictuer

[olive loom]

alright

[olive loom]

ok i feel something is wrong in your steps

this isnt right

how did you get that

[raw bone]

um

,ask d/dy [xy (x^2-y^2)/(x^2+y^2)]

with this

[ocean seal]

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[olive loom]

i see

shall i tell you my way of doing it?

[raw bone]

absolutely

[olive loom]

alright

$f=xy\frac{x²-y²}{x²+y²}$

[ocean seal]

Alpha Kappa

[olive loom]

ok so

the first thing you have to calculate is the partial derivative of f wrt x at (x, 0)

[raw bone]

ok

[olive loom]

at (x, 0) means that you cab substitute y=0

so let me do that

[raw bone]

yes i understand that

[olive loom]

$f=xy\frac{x²-0^2}{x²+0^2}$

[ocean seal]

Alpha Kappa

[raw bone]

what about the first xy?

shouldtn that be x(0)

[olive loom]

oh wait sorry

yeh we are supposed to do that after differentiation

[raw bone]

ok

[olive loom]

ohk now i see why they did it like this

[olive loom]

right then this is correct

[raw bone]

coolio

[olive loom]

but can you do this yourself? like without wolfram?

[raw bone]

umm

...

not really

i kinda understand

but partial differentation is especially rough for me

[olive loom]

hmmm i think you should try yourself

[raw bone]

Havent struggled iwth anything else so far

but this specifically

i suck at

[olive loom]

honestly its simpler than usual differentiation

just take y as a constant and differentiate x

for eg can you do
partial derivate of xy wrt to x?

[raw bone]

yeh

[olive loom]

what will it be

[raw bone]

ok

so the fucntion is xy neh

[olive loom]

yeh

[raw bone]

so its just y

[olive loom]

right

then similarly try partial derivate of the given function in the problem

[raw bone]

okay

but how do i do it in respect af/axay

i dont understanmd how you differentiate wrt both of them

[olive loom]

thats a notation

$\frac{\partial f}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$

[ocean seal]

Alpha Kappa

[olive loom]

it basically means calculate wrt to y first

then take the derivate wrt to x next

[raw bone]

okay so af/ay is what i did earlier on the previous page

but what does it mean if there is only an a/ax

and not af/ax

[olive loom]

look at this i will show you with example

say f=x²y

[raw bone]

okay

[olive loom]

then i will write

$\frac{\partial f}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{\partial (x^2y)}{\partial y}\right)=\frac{\partial}{\partial x}\left(x^2\right)=\frac{\partial x^2}{\partial x}=2x$

[ocean seal]

Alpha Kappa

[olive loom]

can you understand this?

[raw bone]

okay

i understand until it goes ax^2/ax = 2x

only step that doest make sense

at least to me

everything else i understand

[olive loom]

which one

[raw bone]

The (ax^2)/ax = 2x

how does it go to 2x

only thing

also

[olive loom]

thats the usual differentiation no

[raw bone]

oh yeah

that the derivative

[olive loom]

power rule i think its called

[olive loom]

yes

[raw bone]

so does the ax^2/ax

would the ax cancel

leaving x

or is it saying

get the derivative of x^2 wrt x

[olive loom]

this yes

[raw bone]

okay

imma giove this a tryu

[olive loom]

the a is not a variable you cant cancel it
its a function

[raw bone]

be back in l;ike 5 mins

[olive loom]

alright

[raw bone]

can i ping you when done?

or do you not want that \

[olive loom]

yeh sure ping me

[raw bone]

@olive loom

[olive loom]

oh you went ahead and did that 😅

this was not necessary

[raw bone]

fck

oh well

practive i guess

[olive loom]

lol

[raw bone]

when was i supposed to stop

[olive loom]

i just wanted you to try the df/ax and af/ay alone

[olive loom]

you weren’t supposed to do any of this

[raw bone]

is it the right idea

oh

[olive loom]

the first line in this
i just wanted you to do solve that

[olive loom]

how this came

[raw bone]

i was tryna do question 2

rip

[olive loom]

cuz once you get that then b) is super easy

[raw bone]

ok

how was i supposed to do b then

[olive loom]

for b)

[olive loom]

i am taking the df/dy (x, 0) = x and df/dx (0, y) = -y for granted
like i am assuming you proved that

[raw bone]

its givenj

so we know its trye

true

[olive loom]

yep

now you are asked to find df/dxdy right

which means d/dx (df/dy)

substitute df/dy

d/dx (x)

=?

[raw bone]

oh

ok

but the points they used for those values are different

one is x,0 the other is 0,y

[olive loom]

yes the first one is for (x, 0) for all x right
and we need for (0, 0)
so after calculating d(x)/dx you can sub x=0 in it

but it will be a constant anyway

so doesnt matter

what is d(x)/dx = ?

[raw bone]

1

[olive loom]

right

then you have to sub x=0 in it but there is no x so its just 1

thus you get the LHS

df/dxdy at (0, 0) = 1

[raw bone]

okay

[olive loom]

do the same for RHS

[raw bone]

ill try

see you in 5

i wont go fucking differentiate like that again lol

xD

[olive loom]

lol 😂

[raw bone]

This right?

@olive loom

[olive loom]

yeah perfect

[raw bone]

Epic one more.

Question C

[olive loom]

in that question the notation means this:

[raw bone]

just posting again so no need to scroll up

[olive loom]

$\frac{\partial ^2 f}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)$

[ocean seal]

Alpha Kappa

[olive loom]

similarly for y

[raw bone]

okay

so

lemme try

i think i got this

ill send and ping

u know the drill

[olive loom]

alright

[raw bone]

:)

[olive loom]

though wait

@raw bone

[raw bone]

?

[olive loom]

i think the calculation you did was not useless after all

here it does not mention the (x, 0) or (0, y) conditions

[olive loom]

so you will have to do this

which you already did lol

[raw bone]

but i did it for when it d/x(d/y

so it wont be the same

[olive loom]

oh

[raw bone]

also

[olive loom]

right

[raw bone]

its worth 1 mark

which means its gotta be simple

some sort of applicatiopn

wouldnt make sense to be so long for 1 mark

:/

[olive loom]

i see

hmmm

[raw bone]

any ideas?

[olive loom]

thinking

cuz you are right it would be a huge thing

,ask d^2/dx^2 [xy (x^2-y^2)/(x^2+y^2)] + d^2/dy^2 [xy (x^2-y^2)/(x^2+y^2)]

[ocean seal]

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[raw bone]

deosnt make sense for 1 mark

there has to be some sort of thing

cause there is long question here

for ex

This was 5 marks

[olive loom]

then in that case if we take it with the (x, 0) and (0, y) wouldnt it be too simple?

[raw bone]

It is 1 mark :/

[olive loom]

hmmmmmmm

[raw bone]

Cause think this way

[olive loom]

then ig do that itself

[raw bone]

its weird

ill ask around as well

[olive loom]

sure

[raw bone]

thank for all the help alpha

big chad

:)

[olive loom]

np :)

[raw bone]

@olive loom just thought you’d like to see

this isnt right

.close

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[hard garden]

I can't figure out how to do this question

[proud cosmos]

see, AM is normal mean only. So, first do (31/16)x2000. That gives you the sum of all terms of AP till a2000. Then, do 23x(2022-2001+1). This gives sum of all terms from 2001 to 2022. Add both the numbers. Now, divide by 2022. This gives you AM of all terms from 1 to 2022

[hard garden]

Why do we add 1 here (2022-2001+1)

Nvm i got it

Thanks

.close

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[upper pendant]

trying to get the range of this arccos function, when do the signs flip backwards?

[upper pendant]

$0 \le 1-\frac{1}{2}arccos(2x+1)\le \pi$

[ocean seal]

Marevin

[upper pendant]

$1-\frac{1}{2}.0 \le 1-\frac{1}{2}arccos(2x+1)\le 1-\frac{1}{2}\pi$

[ocean seal]

Marevin

[upper pendant]

so it becomes

$1 \le 1-\frac{1}{2}arccos(2x+1)\le 1-\frac{\pi}{2}$

[ocean seal]

Marevin

[upper pendant]

but it is incorrect, the "less than or equal" signs should be flipped, my question is, how and where

[atomic cypress]

when u multiply by the negative half the inequalities flip

[gray ingot]

also you aren't doing the same operation on each side

[upper pendant]

because when you divide or multiply something on an inequality "equation", and the positive or negative signs change, they flip?

am I not?

[gray ingot]

your middle expression is the same

it should also be changed

first you divided by -2 on both sides then added 1

so you have to do that to the 1 - 1/2 arccos(2x+1) too

what you should have done is subtract 1 and multiply by -2

that isolates the arccos(2x+1)

$2 \ge arccos(2x+1) \ge 2-2\pi$

[ocean seal]

Swas.py

[gray ingot]

answering your original question, signs flip when you multiply by a negative value

wait, are you trying to find the range of $f(x)=1-\frac{1}{2}arccos(2x+1)$ using arccos's range?

[ocean seal]

Swas.py

[upper pendant]

I got that answer before but gets me confused because, when I check on a graph like desmos, the range is not [2-2pi; 2]

or co-domain, idk the correct english word

domain is [ -1; 0 ]

[gray ingot]

@upper pendant

[upper pendant]

the range of the function, yes

[gray ingot]

oh

your original inequality was incorrect then

🤦‍♀️

you have to start with
$0 \le arccos(2x+1)\le \pi$

[ocean seal]

Swas.py

[gray ingot]

multiply by -1/2 $0 \ge -1/2arccos(2x+1)\ge -\pi/2$

[ocean seal]

Swas.py

[gray ingot]

add 1, $1 \ge 1-1/2arccos(2x+1)\ge 1 -\pi/2$

[atomic cypress]

^^^

[ocean seal]

Swas.py

[gray ingot]

thats your range

[upper pendant]

Ok, that cleared me up!

Thank you very much to both

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[thick lynx]

You have a 1/100 probability to win a bag.
Every week, you can participate, but it costs 1$.

You have 5$ in total, you can participate multiple times a week, or participate every week for 5 weeks. Which one is better?

[thick lynx]

Is it the same probability?

[prime badge]

yes

[thick lynx]

.close

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[lone heart]

@sonic gate Has your question been resolved?

[kind sierra]

I just want to say, amazing choice of PFP

I literally have a folder full of them

😂

😂

The question is kind of vague, what if the planes intersect?

then the distance would always be 0

[balmy warren]

plug in t = 0

then get difference of vectors

then magnitude

[kind sierra]

How is that the distance between the planes

[balmy warren]

At t= 0

They are vectors

So points in space

Then just find distance between points

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[soft cosmos]

hello

[soft cosmos]

I am having a hard time integrating this. this is literally the only question im having a hard time with

im getting stuck at the u sub part

u=x^2+4

then i got du/dx=2x

then du=2xdx

but i see x^3dx in the question

its really confusing

any help?

[alpine sable]

u=x^2+4 is right and then du/dx=2x so dx=1/(2x) sooooo x^3/2x=1/2(x^2) soooooo x^2/u^(3/4) buuuuut u=x^2+4 therefore x^2=u-4 thus

$\frac{1}{2}\int\frac{u-4}{u^{\frac{3}{4}}}du$

[ocean seal]

Critzzzy

[soft cosmos]

huh?

dx=1/(2x)?

[alpine sable]

sorry forgot the du, dx=1/(2x)du

[soft cosmos]

x^3/2x=1/2(x^2)?

it doesnt make sense hehe

[alpine sable]

$\frac{x^{3}}{2x}=\frac{1}{2}\frac{x^3}{x}=\frac{1}{2}x^2$

[ocean seal]

Critzzzy

[alpine sable]

$\frac{1}{2}\int\frac{u-4}{u^{\frac{3}{4}}}du$

[ocean seal]

Critzzzy

[alpine sable]

forgot to add the 1/2 originally

wanna work through it slower?

[soft cosmos]

well

can you write it in a paper?

i really dont get how you got u^3/4

[alpine sable]

sorry guess I was rushing thats meant to be a 3/2

[soft cosmos]

then integrate it?

[alpine sable]

Yeah and sorry handwriting isn’t the greatest

[soft cosmos]

how did you get 5

[alpine sable]

I mislabeled 6 sorry I'm very sleep deprived. Basically substituted 1/(2x)du for dx

[lone heart]

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[soft cosmos]

@alpine sable thank YOU sir

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[winter zealot]

Hi

[winter zealot]

How can I find the derivative of question i???

My answer is this, but 1 should be 2

[alpine kiln]

yes, that is true

2ln(2x) = 2ln(2)+2ln(x)

the first term in the numerator for the derivative of f(x)/g(x) is g(x)*f'(x)

f'(x)=2/x

then just calculate

[winter zealot]

What is in?

[alpine kiln]

i dont understand your question

[winter zealot]

Like 2 In

2In(2x)

@alpine kiln

[alpine kiln]

i still dont understand

[winter zealot]

This one

[alpine kiln]

can you use more words?

[winter zealot]

I mean I don’t understand this equation

[alpine kiln]

oh i just used

ln(ab) = ln(a) + ln(b)

[winter zealot]

What is In

[alpine kiln]

log_e

log base e of something

[winter zealot]

Ahhhh

Got it got It

[winter zealot]

How does 2loge 2x, the f’(x) became 2/x??

[alpine kiln]

correct

oh i read that wrong

2ln(2) is a constant

so its derivative is 0

ln(x) derivative is 1/x, and so 2ln(x) derivative is 2/x

0+2/x = 2/x

[winter zealot]

I got 1/2x

[alpine kiln]

read over my derivation

if a specific step doesn't make sense

then ask about it

i can't really help you if you just say you got a different answer

[winter zealot]

Ok, I’ll just memorize this

[alpine kiln]

the only thing you need to memorize is the derivative of ln(x) is 1/x

if you multiply it by 2

the derivative also scales by 2

[winter zealot]

Oh

Yes

[alpine kiln]

if you have no more questions do .close

[winter zealot]

.close

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[proud knot]

Determine the equation of the parabola in vertex form.

confused on how we're supposed to get vertex form from the parabola

[proud knot]

vertex is (2,5)

[lone heart]

@proud knot Has your question been resolved?

[proud knot]

i was able to look up a way to find the vertex form but im not sure if this is correct

[limpid spade]

huh

7=a+5
a=2

how u got 12

[proud knot]

7 + 5

its hard to write down equations on google doc hold up

[lone heart]

@proud knot Has your question been resolved?

[proud knot]

should i have kept the + 5? or do i have to switch it to a - to get the answer

[lone heart]

@proud knot Has your question been resolved?

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[dark hull]

is there any further simplification of this function?

[dark hull]

or do i have to create a gigantic circuit

[tacit arch]

Can't be simplified. Polynomials of 7 variables are hard to factor

[prime badge]

what's it mean i'm curious

multi-level?

[dark hull]

multi level gate implementation

[prime badge]

oh i thought it's the gates themselves

it's just nonsense then

[dark hull]

well it's a gigantic circuit then

a pain to draw

[prime badge]

http://csg.csail.mit.edu/6.884/handouts/other/qm.pdf

[dark hull]

what's this

[prime badge]

i don't know, i googled two level dunction

and this seems like super relevant

[placid zinc]

I agree that any random poly of 7 variables might be a nightmare, but they're specifically handing you this one. Maybe there's a nice solution?

[dark hull]

well i found the gate level implementation of it in normal ands and ors

[lone heart]

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[half flax]

how can i parametrize x²+y²=1

[half flax]

sqrt(1-t²)?

nah that makes no sense then it cant be bigger then 1

[alpine sable]

i know the solution

i just dont know how i got there

[half flax]

can i use sin and cos

[alpine sable]

yeah

[half flax]

yeah i thought about that too does it work

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[digital echo]

AB=DC and AB| |DC
prove AD| |BC

please help

[alpine sable]

AD and BC don't look parallel to me

[wary stream]

Do you have an image with this?

[digital echo]

yes i might of written it incorrectly

i dont know the answers to the other questions eihter

[frozen python]

just use the properties of the parallelogram

[digital echo]

i cant it doesnt say its a parallelogram

[frozen python]

AB=DC and AB||DC =>ABCD parallelogram

@digital echo

[digital echo]

<@&286206848099549185> is it correct?

[lone heart]

@digital echo Has your question been resolved?

[digital echo]

need help plss

[clear stump]

$x+20=2x-60$

[ocean seal]

.,..

[clear stump]

@digital echo

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[lone heart]

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[alpine sable]

Where you stuck

[wintry silo]

tbh i dont know where to start

[alpine sable]

Do you know I substitution?

U sub

[wintry silo]

yes

[wary stream]

What substitution should you use?

[weary wyvern]

Try a random substitution

Use ur intuition

[wintry silo]

ok thx

[wintry silo]

the u would be u=x^2 right?

[tender dew]

that's my first thought

[alpine sable]

Yeah so what next?

[tender dew]

but then solving by parts won't either erase nor diminish sqrt(x)

mvm

consequence du should do the trick

[wintry silo]

du=2x

[tender dew]

integrate (1/2)*sin(u)du

[alpine sable]

You’re taking a derivative of u with respect to x so you should have du/dx=2x and then solve for dx since you wanna sub du and it’s addons in for dx

[wintry silo]

ok is the answer -cos(x^2)/2 + C

[alpine sable]

Where’s the /x come from and where’s the 1/2 go

[wintry silo]

sorry i meant 2

not x

[alpine sable]

Then yes it is correct

[wintry silo]

ok thank you for the help lol im kinda slow when it comes to calc

[alpine sable]

It’s fine it’s a hard subject just keep studying you’ll do great

[wintry silo]

ok thnx

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[open roost]

3cosΘ=cosΘ+1

[open roost]

how would u solve this in trig?

[wary stream]

Get all the terms of theta on one side

Then isolate that term

[open roost]

3cosΘ=cosΘ+1

2cosΘ=1

like something like that?

[wary stream]

Yes

Now how can you isolate cos theta?

[open roost]

yea

wait but how does 3cos turn to 2cos?

[wary stream]

If you did 3x = x + 1, do you think you can solve for x?

[open roost]

oh so we think of it like that

like we think cos as x

r?

[wary stream]

You can, yes

[open roost]

ah

[wary stream]

In the end, you just have to remember to put back in cos theta, to solve for theta

[open roost]

the third and eighth terms of an arithmetic sequence are 12 and -18, respectively; determine the fifth term of the sequence

[open roost]

k

could u also help me with this q?

like for this one my first thing was to try systems

[wary stream]

Sequences were not my strong point, sorry

[open roost]

ah k its good

[tender dew]

if it's arithmetic then you know it's linear

[open roost]

we use systems for this r?

[tender dew]

so from 3 to 8 it has 5 differences right?

[open roost]

like
12 = t1 + (3-1)d
-18 = t1 + (8-1)d

?

wait so something like this?

[tender dew]

5 equal parts to boost it from 12 to -18

[open roost]

k

[tender dew]

not sure where -1 came from

but rest seems fine

[open roost]

oh its from the formula

[tender dew]

oh I know

[open roost]

tn = t1 + (n-1)d

[tender dew]

It's for proper indexation

so for 1 it will be just t1

ok

[open roost]

k

[tender dew]

answer should be 0

[open roost]

k let me try

oh yea it is

thx so much man

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[restive hollow]

Can somebody help me with 30) please

[lone heart]

@restive hollow Has your question been resolved?

[restive hollow]

Okay nvm with that.. can somebody help me tho with knowing how to draw angles from words?

For example “25 degree angle to the vertical”

Like how do I know what that looks like

.close

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[narrow compass]

hello i wanted to ask how and where to get this number the one that i circled (red)

[alpine sable]

theres 4 circled numbers

[narrow compass]

the red one

[alpine sable]

there getting all the denominators to 20 so the can add them together

[narrow compass]

ohh thank you i didnt notice that. i wasnt listening well on my teacher and ws to afraid to ask :3 but thank you

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[grave robin]

Why is the rearrangement (-2e^x + 3) rather than (2e^x - 3)? its been a while since I last had to rearrange equations

[tender dew]

same equation

if you equate to 0

[grave robin]

yeah but I did the rearranging by moving the 3 first then the other e and I got a different answer when I found the area

[tender dew]

ok

whixh function you think is on top

?

[grave robin]

uh no idea

im not entirely sure I calculated it correctly

[tender dew]

seriously?

[grave robin]

ive not done this type of q before

This is what I got

[tender dew]

you gotta plot those

draw or use wolfram alpha

[grave robin]

I'd rather not have to use those so I know how to do it for an exam situation

[tender dew]

then you will know if it's either f(x)-g(x) or g(x)-f(x)

you know how to integrate but not know how the function looks like?

[grave robin]

haha yeah, its been a while since I've done any math like this and we just did integration last week

[tender dew]

ok simplier

just use absolute value

of you are not bound by the integral notion

since |-2e^x +3| = |2e^x -3|

beware

for if you are

for negative fields

then you'd need to evaluate

[grave robin]

okay, I'll have to find out if I need to do it a certain way. Thank you for your help

[tender dew]

with the equations you'd shown

it looks like you ARE

[grave robin]

ah ok, ty

[tender dew]

no wait

it has even more complex answer

those function intersect

and if you want to be given just positive are

then you need to divide and then integrate specific parts

like here

first you fpund where they do and that it is within searched boundary

then you figured in first part f(x) is on top and in second g(x) is on top

[grave robin]

that just sounds complicated :/

i think my different answer is coming from the ln he used but I'm not sure why that was done

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[heady igloo]

someone help pls

[alpine sable]

Read the proof carefully and think about it. Everything you need is already given

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[lone heart]

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[alpine sable]

alr thank you

i got 78.4°?

thank you

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[uneven orbit]

6^x+2=3^x. Enter an exact answer or round your answer to the nearest tenth.

my answer x=-5.1

[vale wigeon]

do you mean $6^{x+2}=3^x$ or $6^x + 2 = 3^x$?

[ocean seal]

Ann

[uneven orbit]

The former

The final answer I got was −5.169922 but since I rounded to the nearest tenth it should be -5.1 no?

$6^{x+2}=3^x$

[ocean seal]

Cantiflas

[austere garnet]

That rounds to -5.2

[uneven orbit]

welp

.closed

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[merry depot]

did you honestly just ping 11000 people to get help with something you can google?

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[stable spire]

??

[austere garnet]

The ping didn't go through since they don't have perms

Still just google it though

[stable spire]

.close

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[lone heart]

@alpine sable Has your question been resolved?

[thick yoke]

First you need to find the line's equation.

Then, find the intersection point of the line and the y axis,

and then, calculate the distances and divide the first by the second to find the ratio.

[oak perch]

Or he can calculate ratio of distance between two points and line x=0 respectively

Which is |3|:|1|

[alpine sable]

yeah that's what I tried to do but I can't find the distance between the points and x=0

[oak perch]

?
Just |their x value| respectively

[alpine sable]

wait really

the line segment joining the points and x=0 should coincide with the line segment joining the points right

are we talking about the straight lines or the slanted ones

@oak perch

[oak perch]

Length of Two horizontal line segments in your picture

[alpine sable]

but they would be shorter than the actual distances right

[oak perch]

The ratio is the same

[alpine sable]

why

[oak perch]

Two similar triangles in your picture

[alpine sable]

ah ok i see

they're like corresponding sides

[oak perch]

yeah

[alpine sable]

so would all the side lengths of the triangle have a ratio of 3:1

[oak perch]

Yeah. And Whenever you have three points p_j=(x_j,y_j,z_j,…) on the same line
d(p_1,p_2):d(p_2,p_3)=|x_2-x_1|:|x_3-x_2|=|y_2-y_1|:|y_3-y_2|=|z_2-z_1|:|z_3-z_2|=…

[alpine sable]

ok

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[clear stump]

That physics? @alpine sable

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[safe lark]

could someone help me with this, solutions say 25cm but i get 25/11 cm
bisector of angle beta in a triange ABC "cuts" the side AC in point D. A parralel DE with AB, E is element of BC, passes through point D. Then a parallel EF with BD, F is element of AC passes through point E. What is the length of side b if a=30cm, c=20cm, |AD|-|CF|=1cm

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@safe lark Has your question been resolved?

[safe lark]

<@&286206848099549185>

[safe lark]

<@&286206848099549185> please someone

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@safe lark Has your question been resolved?

[safe lark]

<@&286206848099549185>

[robust glacier]

Calm down bro

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@safe lark Has your question been resolved?

[safe lark]

<@&286206848099549185>

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@safe lark Has your question been resolved?

[alpine sable]

Hey

[safe lark]

hey

[safe lark]

would you know what's wrong, please i've been asking for help everywhere the whole day

[noble salmon]

I think I can help you, can you label out each side with "a,b,c", so that it'll be easier to visualise?

[safe lark]

a would be the one opposite of point A so BC, b is CA, c is AB

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[simple turtle]

@alpine sable

[simple turtle]

It looks sort of like this, isn't it?
You have to play around little bit, but you can also use the formula shown here. Of course if that is what you are seeking. https://www.symbolab.com/solver/functions-graphing-calculator/graph x %3D -\frac{1}{500}\left(y - 9000\right)^{2} %2B 5000?or=input

Your thought?

that is great.

what formula did you use?

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[low tartan]

Hi

[low tartan]

How do I find an “equation for the axis of symmetry” for q(x)=x^2+10x+16

[languid portal]

Graphing the equation, we can see it more clearly

at what x value would you say the parabola is reflected across?

[gray ingot]

find the vertex

then the x = x cord of the vertex is the line of symmetry

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@low tartan Has your question been resolved?

[low tartan]

How do I make it transform 3 to the left

[native hull]

Think about the order of operations

[low tartan]

What do you mean

I know changing up and down is the number at the end

But I don’t know for horizontal

[native hull]

Test something things on a graphing tool

[low tartan]

I already tried

[native hull]

If I remember correctly you have to change the form of the equation

[alpine sable]

yoyy

yototo

ıddodo

yoyoyo

[native hull]

Change it into the form a(x+b)^2+c

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[cloud rain]

Can someone help me pls

[cloud rain]

the answer to a is 5
and b)i is 20.8

I'm stuck at b)ii

also b)iii

[dusty rivet]

For b)ii
You can find the length of CD using either Pythagorean theorem or 45-45-90 triangles

Then, find the altitude of CD (which would be the height of triangle BCD

Finally, area formula of a triangle

I think that should work

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[small token]

(sin/csc)+(cos/sec) - (cos) (sec) = 0

[small token]

Hello I need a little help with some notes I know the identities just need a little push in the right direction

nvm I figured it out 😄

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[jaunty silo]

I need help with the second part of the first one

And b c and d on the other one

I think that’s right

Just c,d and the second part of the first one

<@&286206848099549185>

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[sand star]

hi guys , i know it is to integrate but i am still kinda confuse cause the graph is looking really weird and i am not sure what should i be finding

[stuck grail]

The area enclosed between the two points where the red and blue curves intersect

[limpid spade]

solve both equations for y youll get y=+-sqrt(something) + and - one are the same but let's say rotated about the x-axis where the x-axis is the symmetrical axis, then find the intersection points and multiply the integral by two

[sand star]

wait i dont really get it , so first i want to integrate them serperature which result in two different equation

then i use one minus another?

[oak perch]

Like I said, integral of (49/25-y^2)dy y from -7/5 to 7/5

[stuck grail]

Err, you can easily integrate this with respect to dy, no need to solve for y.

[limpid spade]

ahhh hahaha

[sand star]

why is it y^2?

[oak perch]

?

Your two functions

49/25-y^2/2

And y^2/2

The difference is 49/25-y^2

[sand star]

ohhh alright tysm!

btw how do you identify which functoin minus by another

[sand star]

like why cant it be function 2 - function 1?

[oak perch]

?

y^2/2 is below

The other one is above

[sand star]

ohh like without a calculator or graph, how will you know that

cause it is not suppose to use a graphing calculator

[limpid spade]

plug in values and see which one has higher x values in this case

[oak perch]

?

You can integral 49/25-y^2?

(49/25)y-y^3/3

The difference of it at value 7/5 and -7/5 respectively

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@sand star Has your question been resolved?

[lone heart]

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[alpine sable]

Q5iii

[supple tundra]

What have u tried to do

[alpine sable]

I tried using roots of polynomials way seems too long

I’ll be honest I got no clue. Done part one and two but don’t know how to do part 3

a=-32

[supple tundra]

If I gave you this $x^2+ax+b=0$

[ocean seal]

azeem321

[supple tundra]

can you solve

for x

[alpine sable]

b=1156

Yeah

[supple tundra]

So do that and then just let $y^2=x$

[ocean seal]

azeem321

[alpine sable]

Yh but the hidden quadratic method

I tried that

[supple tundra]

hmm

[alpine sable]

And I just get y^2 = (16+/-30i)

So I end up knowing something I already knew

[supple tundra]

How do you find square root of a complex number?

So we have $y^2=16+30i$ and $y^2=16-30i$

[ocean seal]

azeem321

[alpine sable]

Yh

[supple tundra]

What is $\sqrt{16+30i}$

[ocean seal]

azeem321

[alpine sable]

Idk

You can’t simplify that any further surely

[supple tundra]

Let's suppose that $\sqrt{16+30i} = a+bi \iff 16+30i=(a+bi)^2 = a^2-b^2+2abi$

[ocean seal]

azeem321

[supple tundra]

So we have $2ab=30$ and $a^2-b^2=16$

[ocean seal]

azeem321

[supple tundra]

can you solve these? for a,b

[alpine sable]

Oh shi okay that makes sense thanks

[supple tundra]

is this a calculator paper

[alpine sable]

Yh

[supple tundra]

then you could've turned to polar/exponential form

[gray isle]

though don't use a,b as those are already established variables in the question

[supple tundra]

ohh

whoops

sorry

[alpine sable]

Yh I’ll use p and q

Anyway cheers

.close

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[ornate grotto]

how would i go about using law of sines to find out length AB

[supple tundra]

What did you learn in law of sines

[wary stream]

Do you know law of sines?

[ornate grotto]

yes, sin(A)/a=sin(B)/b=sin(C)/c

[wary stream]

Can you apply that formula with two angles and it's corresponding side?

[ornate grotto]

so what i get is

i already did this step but i was unsure of how to simplify it now

(I know that sin(30) = 1/2)

[wary stream]

You don't have to have that middle fraction

[ornate grotto]

Oh really?

so it just gives us

and then I can solve from c here easily

thank you very much

.close

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[wary stream]

Because there's an equal sign

[ornate grotto]

you're right

[wary stream]

So A = B = C, means that A = C

[ornate grotto]

I understand now, thank you

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[lost coral]

am i right?

[wary stream]

You can check using a calculator

Type in 5^4 * 5^1 and compare that with 5^5

[lost coral]

hmm is this correct?

[covert agate]

yea

[lost coral]

hmm how do i do this

[covert agate]

what have you tried

[lost coral]

like this?

[covert agate]

good

[lost coral]

cause if its a large number it will take a long time to find it

[lost coral]

is there a way to find 4^3 other than testing it

[wary stream]

If it's no calculator, mostly your teacher would use small numbers

[lost coral]

something like this but power 4

[lost coral]

nvm i found a way

just use the factors of 4

[wary stream]

Or just know your cubes, up to like 6. That's all I know up to

[lost coral]

wait am i correct?

reciprocal of 1/3 is 3/1

so its 3?

[supple tundra]

yh

[dense hazel]

yes

[lost coral]

then why did google lied to me

[wary stream]

Like 5^3 = 125 or 6^3 = 216

[supple tundra]

what did google say

[lost coral]

oh ok

ill memorize cube root numbers

and square roots as well

[wary stream]

Because you typed it in wrong

[lost coral]

oh i've forgotten to add parentheses

[wary stream]

It took the reciprocal of just the numerator

[lost coral]

oo

i have problems

[wary stream]

Apply exponent properties

[lost coral]

this is my attempt

[wary stream]

What's 10/4 simplified?

[lost coral]

oh yea

[wary stream]

Yes

[lost coral]

oh wait

the r is supposed to be at the top

[wary stream]

It is

[lost coral]

where do i put the q?

numerator or the deminator?

[wary stream]

What's q^-1 mean?

[lost coral]

1/q

[wary stream]

So is that numerator or denominator?

[lost coral]

denominator

so if we apply that to a positive power

then it would be at the top?

[wary stream]

Yes

[lost coral]

thank you

wait so is it q^-1 or just q?

[wary stream]

.

Apply that

[covert agate]

typically we express the answer with positive indices

[wary stream]

You already said 1/q

Which is 1/q^1

[lost coral]

so we are trying to remove it

so we can write it like this?

its the same but we are just removing the negative

[wary stream]

You can do that or apply the negative exponent concept

[lost coral]

what is the exponent is like -6?

so we write it as q^6?

just removing the negative sign

[wary stream]

Where $a^{-1} = \frac{1}{a}$

[ocean seal]

dldh06

[lost coral]

ok

[lost coral]

hey dldh06 i need some help with this

for that top question

[wary stream]

Please do not ping me

[lost coral]

oh ok

like this?

[wary stream]

Looks fine

[lost coral]

ok

how?

[lilac stratus]

Factor both the numerator and denominator and see if you can cancel anything out.

[lost coral]

heres my attempt

[lilac stratus]

You can't just get rid of those two -4s.

[lost coral]

but i cant factor both if -4 doesnt has x

unless i add x two each terms

and make it into a cubic

give me a sec

like this?

and i do this

[lilac stratus]

That last part you'll have to do, yes. But there's no need to multiply by x first.

You can make both the numerator and denominator equal to 0, solve both equations and factor from there.

[lost coral]

oh let me try

[lost coral]

its not working

gotcha

i found it

u mean this?

3x-1 is the top

[lilac stratus]

Exaclty.

[lost coral]

but what do i do with that -4?

do i just forget about that -4?

[lilac stratus]

No. The original polynomial ($3x^2+11x-4$) will be equal to the product of those two things you have there (so $(x+4)(3x-1)$).

[ocean seal]

JaimermXD

[lost coral]

oh i see

so i dont forget about that -4